These problems draw on the whole book and are not labelled by chapter. For each, decide first what kind of question is being asked and which model applies; that decision is part of the problem.

1. The numbers of flaws in \(12\) rolls of fabric total \(30\) , and the counts are modelled as a random sample from a Poisson distribution with mean \(\lambda\) . (a) Give the maximum likelihood estimate of \(\lambda\) , its approximate standard error, and an approximate \(95\%\) confidence interval. (b) Estimate the probability that a roll has no flaws by maximum likelihood and by the best unbiased estimator, and explain why the two differ.

   Solution

   (a) For a Poisson sample the log-likelihood is \(30\log\lambda-12\lambda\) up to a constant, maximised at \(\hat\lambda=30/12=2.5\) . The Fisher information in the sample is \(12/\lambda\) , so the approximate standard error is \(\sqrt{\hat\lambda/12}=\sqrt{2.5/12}\approx0.456\) , and the Wald interval is \(2.5\pm1.960(0.456)=(1.61,\ 3.39)\) .

   (b) By invariance the MLE of \(e^{-\lambda}\) is \(e^{-2.5}\approx0.0821\) . The total \(T=30\) is complete and sufficient, and conditioning the indicator that the first roll is flawless on \(T\) gives the UMVUE \((1-1/12)^{30}=(11/12)^{30}\approx0.0735\) , as in Example 6.2.

   The two differ because \(e^{-\lambda}\) is a convex function of \(\lambda\) : substituting an unbiased estimate of \(\lambda\) into a convex function overestimates on average, and the unbiased estimator is smaller to compensate. With \(n=12\) the difference is about \(10\%\) of the estimate; it vanishes as \(n\) grows, since \((1-1/n)^{n\bar x}\to e^{-\bar x}\) . Neither is “the” answer: the MLE is the plug-in value that is efficient in large samples, and the UMVUE is correct on average at this sample size.

2. Six observations from the uniform distribution on \((0,\theta)\) have maximum \(7.2\) . Give the maximum likelihood estimate of \(\theta\) , the best unbiased estimate, the shortest \(95\%\) confidence interval based on the maximum, and the p-value for testing \(H_0:\theta\ge10\) against \(H_1:\theta<10\) .

   Solution

   The likelihood is \(\theta^{-6}\) for \(\theta\ge7.2\) and zero below, so the MLE is \(\hat\theta=7.2\) . The maximum \(M\) is complete and sufficient with \(E\,M=\frac67\theta\) , so the UMVUE is \(\frac76(7.2)=8.4\) .

   For the interval, \(M/\theta\) has distribution function \(u^6\) on \((0,1)\) , and by Exercise 7.6 the shortest interval from this pivot is \([M,\ M\alpha^{-1/6}]\) . With \(\alpha=0.05\) , \(0.05^{-1/6}\approx1.6475\) , giving \([7.2,\ 11.86]\) .

   For the test, small values of \(M\) are evidence against \(H_0:\theta\ge10\) . The p-value is the supremum over the null hypothesis of \(P_\theta(M\le7.2)=(7.2/\theta)^6\) , which is decreasing in \(\theta\) and so attained at the boundary \(\theta=10\) :

   \[ p=0.72^6\approx0.139 . \]

   The hypothesis \(\theta\ge10\) is not rejected at the \(5\%\) level. This agrees with the confidence interval, which extends to \(11.86\) and therefore contains values of \(\theta\) above \(10\) ; the one-sided test and the interval are two readings of the same pivot.

3. An instrument records only the sum \(Y=A+B\) of two independent quantities, \(A\sim N(\mu_A,1)\) and \(B\sim N(\mu_B,1)\) . Explain what can and cannot be estimated consistently from repeated observations of \(Y\) , prove your claim, and describe one change to the data collection that would make \((\mu_A,\mu_B)\) estimable.

   Solution

   \(Y\sim N(\mu_A+\mu_B,\ 2)\) . The distribution of the data depends on \((\mu_A,\mu_B)\) only through the sum \(s=\mu_A+\mu_B\) .

   What can be estimated. The sum is identifiable, being the mean of \(Y\) , and \(\bar Y\) is a consistent estimator of it by the Weak Law.

   What cannot. The parameter values \((\mu_A,\mu_B)\) and \((\mu_A+c,\ \mu_B-c)\) give the same distribution of every sample, for any \(c\ne0\) , while \(\mu_A\) differs between them. By Theorem 1.9, no sequence of estimators is consistent for \(\mu_A\) , nor for \(\mu_B\) , nor for \(\mu_A-\mu_B\) . The proof of that theorem applies verbatim: an estimator that concentrated near \(\mu_A\) under the first parameter value would, having the same distribution under the second, concentrate near \(\mu_A\) there too, and so not near \(\mu_A+c\) .

   A repair. Change the design so that the data distinguish the two components: for instance, on some occasions disconnect the second source and record \(A\) alone. Then \(\mu_A\) is the mean of those observations and \(\mu_B\) is the mean of the sum minus \(\mu_A\) . Alternatively, record \(A-B\) as well as \(A+B\) . Collecting more observations of the sum alone would not help at all.

4. Eight lifetimes from an exponential distribution total \(40\) hours. Estimate the probability that a lifetime exceeds \(10\) hours by maximum likelihood, and derive the best unbiased estimator by conditioning the indicator that the first lifetime exceeds \(10\) on the total, using the fact that \(X_1/\sum X_i\) has density \((n-1)(1-u)^{n-2}\) on \((0,1)\) and is independent of the total.

   Solution

   Maximum likelihood. \(\hat\lambda=8/40=0.2\) per hour, and by invariance the MLE of \(P(X>10)=e^{-10\lambda}\) is \(e^{-2}\approx0.1353\) .

   Best unbiased estimator. The total \(T=\sum X_i\) is complete and sufficient. Start from the unbiased estimator \(U=\mathbf 1\{X_1>10\}\) and condition on \(T\) . The ratio \(W=X_1/T\) is unchanged by rescaling all the lifetimes, so it is ancillary, and by Basu’s theorem it is independent of \(T\) . Hence, for \(t>10\) ,

   \[ P(X_1>10\mid T=t)=P\bigl(W>10/t\bigr)=\int_{10/t}^1(n-1)(1-u)^{n-2}\,du=\Bigl(1-\frac{10}{t}\Bigr)^{n-1}, \]

   and the conditional probability is \(0\) when \(t\le10\) . By the Rao–Blackwell and Lehmann–Scheffé theorems this function of \(T\) is the UMVUE. With \(n=8\) and \(t=40\) ,

   \[ \Bigl(1-\frac{10}{40}\Bigr)^7=0.75^7\approx0.1335 . \]

   The two estimates, \(0.1353\) and \(0.1335\) , are close. As \(n\to\infty\) with \(t/n\) fixed, \((1-c/t)^{n-1}\to e^{-cn/t}\) , the MLE. The unbiased estimator has the sensible property of being exactly zero when the total time on test is less than \(10\) hours, in which case no single lifetime could have exceeded \(10\) .

5. A supplier claims a defect rate below \(5\%\) . In \(200\) items, \(4\) are defective. Set up hypotheses that place the burden of proof on the supplier, compute the exact p-value, determine the rejection region of the level-\(0.05\) test based on the count, and find its power when the true rate is \(2\%\) .

   Solution

   The claim to be established is \(p<0.05\) , so it is the alternative: \(H_0:p\ge0.05\) against \(H_1:p<0.05\) . The supplier must produce evidence; absence of evidence leaves \(H_0\) standing.

   With \(X\sim\operatorname{Binomial}(200,p)\) , small counts support \(H_1\) . The binomial family has monotone likelihood ratio, so the supremum of \(P_p(X\le x)\) over \(p\ge0.05\) is at \(p=0.05\) , and the p-value is

   \[ P_{0.05}(X\le4)\approx0.0264 . \]

   For the level-\(0.05\) test: \(P_{0.05}(X\le4)=0.0264\) and \(P_{0.05}(X\le5)=0.0623\) , so the rejection region is \(X\le4\) , with actual size \(0.0264\) . The observed count \(4\) lies in it: reject \(H_0\) and accept the claim at the \(5\%\) level.

   Power at \(p=0.02\) : \(P_{0.02}(X\le4)\approx0.629\) . Even if the true rate is \(2\%\) , well under the claimed bound, a sample of \(200\) would establish the claim only \(63\%\) of the time. Had \(5\) defectives been found, with \(\hat p=0.025\) , half the claimed bound, the test would not have rejected. A sample of this size can confirm the claim only when the defect rate is very low.

6. For the data of Problem 5, a buyer holds the prior \(\operatorname{Beta}(1,19)\) for the defect rate. Find the posterior distribution, its mean, and the posterior probability that the rate is below \(5\%\) . Explain why this probability and the p-value of Problem 5 can both be reported without contradiction.

   Solution

   By Proposition 11.2(a) the posterior is \(\operatorname{Beta}(1+4,\ 19+196)=\operatorname{Beta}(5,215)\) . Its mean is \(5/220\approx0.0227\) , between the prior mean \(0.05\) and the sample proportion \(0.02\) and much closer to the latter, since the prior is worth \(20\) observations against \(200\) . The posterior probability that the rate is below \(5\%\) is

   \[ P(p<0.05\mid x)=P\bigl(\operatorname{Beta}(5,215)<0.05\bigr)\approx0.986 . \]

   The p-value of Problem 5, \(0.026\) , is the probability of observing four or fewer defectives if the rate were exactly \(5\%\) . The posterior probability \(0.986\) is the probability that the rate is below \(5\%\) given the data and the prior. They are probabilities of different events computed under different assumptions: the first conditions on a parameter value and describes data, the second conditions on the data and describes the parameter. That \(0.026\) and \(1-0.986=0.014\) are of similar size is common for one-sided problems with weak priors, and it is not an identity. Reporting both is consistent as long as each is labelled for what it is; the error would be to describe the p-value as the probability that the supplier’s claim is false.

7. The points \((x,y)\) are \((0,1.1)\) , \((1,1.9)\) , \((2,3.2)\) , \((3,3.8)\) , \((4,5.0)\) . Fit a straight line by least squares, estimate \(\sigma^2\) , test the hypothesis that the slope equals \(1\) , and give a \(95\%\) prediction interval for a new observation at \(x=5\) . State one feature of the residuals you would examine before trusting the interval.

   Solution

   Here \(n=5\) , \(\bar x=2\) , \(\bar y=3.0\) , \(S_{xx}=10\) , and \(S_{xy}=\sum(x_i-2)y_i=-2.2-1.9+0+3.8+10.0=9.7\) . So \(\hat\beta_1=0.97\) and \(\hat\beta_0=3.0-0.97(2)=1.06\) .

   The fitted values are \(1.06,2.03,3.00,3.97,4.94\) and the residuals \(0.04,-0.13,0.20,-0.17,0.06\) , with \(\mathrm{RSS}=0.0910\) and \(\hat\sigma^2=0.0910/3\approx0.0303\) .

   For \(H_0:\beta_1=1\) : the standard error of the slope is \(\sqrt{0.0303/10}\approx0.0551\) , so

   \[ t=\frac{0.97-1}{0.0551}\approx-0.54 \]

   on \(3\) degrees of freedom, with two-sided p-value about \(0.62\) . A slope of \(1\) is entirely compatible with the data.

   Prediction at \(x_0=5\) : the point forecast is \(1.06+0.97(5)=5.91\) , with standard error \(\sqrt{0.0303\,(1+\tfrac15+\tfrac{9}{10})}\approx0.252\) . With \(t_{3,0.025}\approx3.182\) the interval is \(5.91\pm0.80=(5.11,\ 6.71)\) .

   The point \(x=5\) lies outside the observed range, so the interval depends on the line continuing to hold. Before trusting it, plot the residuals against \(x\) and look for curvature: a systematic pattern such as positive, negative, positive would mean the mean function is not a line, and the extrapolation would then be biased by an amount the interval does not include. With five points the check is weak, which is itself a reason for caution.

8. A report says: “The difference between the two groups was significant, \(p=0.04\) , so there is a \(96\%\) probability that the treatment works; the \(95\%\) confidence interval for the mean difference is \((0.1,\ 2.3)\) , so \(95\%\) of patients improve by between \(0.1\) and \(2.3\) units.” Identify each error and rewrite the two conclusions correctly.

   Solution

   First sentence. A p-value of \(0.04\) means that, if the treatment had no effect, a difference at least as large as the one observed would arise in \(4\%\) of studies. It is computed assuming the null hypothesis and is not the probability that the null hypothesis is true; “\(96\%\) probability that the treatment works” would be a posterior probability, which requires a prior and a different calculation. Correct version: “The observed difference would be unusual if there were no effect (\(p=0.04\) ); the data provide moderate evidence of a difference.”

   Second sentence. The confidence interval \((0.1,2.3)\) concerns the mean difference between groups, a parameter. It says nothing about the distribution of individual patients’ responses, which is far wider than the uncertainty about the mean, and “improve by” attributes to each patient an individual effect that the study cannot observe. Nor does \(95\%\) describe a proportion of patients: it is the long-run fraction of such intervals that contain the true mean difference. Correct version: “We estimate that the treatment raises the mean outcome by between \(0.1\) and \(2.3\) units (\(95\%\) confidence interval); individual responses vary much more widely.”

   The corrected report also reveals what the original concealed: the lower limit \(0.1\) is close to zero, so the effect, while probably real, may be small.

9. For the Poisson data of Problem 1, obtain a \(95\%\) confidence interval for the probability \(e^{-\lambda}\) of a flawless roll in two ways: by transforming the endpoints of the interval for \(\lambda\) , and by the delta method applied directly to \(e^{-\hat\lambda}\) . Compare the two intervals and say which you would report.

   Solution

   Transformation. The function \(\lambda\mapsto e^{-\lambda}\) is strictly decreasing, so the event that \(\lambda\) lies in \((1.605,\ 3.395)\) is the event that \(e^{-\lambda}\) lies in \((e^{-3.395},\ e^{-1.605})=(0.034,\ 0.201)\) , and the coverage is the same.

   Delta method. With \(g(\lambda)=e^{-\lambda}\) , the standard error of \(e^{-\hat\lambda}\) is \(e^{-2.5}\times0.456\approx0.0375\) , and the interval is \(0.0821\pm1.960(0.0375)=(0.009,\ 0.156)\) .

   The two differ substantially. The transformed interval is asymmetric about the estimate, extending further upward, and it stays inside \((0,1)\) whatever the data. The delta-method interval is symmetric, comes within \(0.009\) of zero here, and for smaller samples would include negative values. Both rest on the same normal approximation for \(\hat\lambda\) , applied on different scales, and the approximation is better on the scale of \(\lambda\) , where the sampling distribution of \(\hat\lambda\) is closer to normal, than on the scale of \(e^{-\lambda}\) , where it is strongly skewed. I would report the transformed interval. Better still would be to transform the endpoints of an interval for \(\lambda\) that is itself more accurate, such as the square-root interval of Example 9.3 or the exact interval.

10. In a matched-pairs experiment with random assignment within pairs, the five differences, treated minus control, are \(3\) , \(1\) , \(4\) , \(2\) , \(-1\) . Compute the one-sided randomization p-value for the mean difference by counting sign patterns, compute the paired \(t\) statistic and its two-sided p-value, and say what each analysis assumes.

    Solution

    Randomization test. Under the sharp null hypothesis each difference is equally likely to have either sign, giving \(2^5=32\) equally likely sign patterns for the magnitudes \(3,1,4,2,1\) . The observed sum is \(9\) and the largest possible is \(11\) . A pattern has sum at least \(9\) when the magnitudes given a negative sign total at most \(1\) : no negatives, or a single negative on one of the two magnitudes equal to \(1\) . That is \(3\) patterns, so

    \[ p=\frac{3}{32}\approx0.094 . \]

    Paired \(t\) . The mean is \(\bar d=1.8\) ; the deviations \(1.2,-0.8,2.2,0.2,-2.8\) have squares summing to \(14.8\) , so \(s_D^2=3.7\) and \(s_D\approx1.924\) . The standard error is \(1.924/\sqrt5\approx0.860\) and \(t=1.8/0.860\approx2.09\) on \(4\) degrees of freedom, with two-sided p-value about \(0.105\) , hence one-sided about \(0.052\) .

    Assumptions. The randomization test assumes only that a fair coin assigned treatment within each pair; it tests the sharp null that no unit is affected, and it is exact. The \(t\) -test assumes the differences are a random sample from a normal distribution; it tests that the mean difference is zero. Neither analysis finds the evidence strong, which is what five pairs with one reversal can be expected to show. The smallest p-value any five-pair experiment could yield from the randomization test is \(1/32\approx0.031\) .

11. Two forecasters gave probabilities for five events. Forecaster A gave \(0.9\) , \(0.8\) , \(0.7\) , \(0.2\) , \(0.1\) and forecaster B gave \(1.0\) , \(0.9\) , \(0.5\) , \(0.4\) , \(0.0\) ; the first two events occurred and the last three did not. (a) Compute each forecaster’s mean Brier score and mean logarithmic score. (b) Recompute both scores supposing that the fifth event had occurred as well, and explain what the comparison shows about forecasts of \(0\) and \(1\) .

    Solution

    (a) Outcomes \(1,1,0,0,0\) .

    Brier, A: \(\tfrac15(0.01+0.04+0.49+0.04+0.01)=0.118\) . Brier, B: \(\tfrac15(0+0.01+0.25+0.16+0)=0.084\) .

    Logarithmic, A: \(-\tfrac15(\log0.9+\log0.8+\log0.3+\log0.8+\log0.9)\approx0.372\) . Logarithmic, B: \(-\tfrac15(\log1+\log0.9+\log0.5+\log0.6+\log1)\approx0.262\) .

    Forecaster B does better on both scores. B’s two categorical forecasts, \(1.0\) and \(0.0\) , were both right and cost nothing.

    (b) Outcomes \(1,1,0,0,1\) . Only the fifth term changes. Brier: A’s fifth term becomes \(0.81\) and the mean \(0.278\) ; B’s becomes \(1\) and the mean \(0.284\) . Logarithmic: A’s fifth term becomes \(-\log0.1\approx2.30\) and the mean about \(0.812\) ; B’s becomes \(-\log0\) , which is infinite, so B’s mean score is infinite.

    One event reverses the ranking under the Brier score and, under the logarithmic score, ruins forecaster B beyond recovery: no number of good forecasts afterwards brings an infinite total back. A forecast of exactly \(0\) or \(1\) is a claim of certainty. The Brier score charges a bounded price, at most \(1\) , when the claim fails. The logarithmic score charges an unbounded one, on the reasoning that an event declared impossible has likelihood zero. Unless the forecaster really has certainty, the rational report under either rule is strictly between \(0\) and \(1\) .

12. Two laboratories measure the same quantity \(\mu\) . The first reports the mean \(10.4\) of \(4\) measurements with standard deviation \(1\) each; the second reports the mean \(9.8\) of \(16\) measurements with standard deviation \(2\) each; all errors are normal with known standard deviations. Find the maximum likelihood estimate of \(\mu\) from the combined data, a \(95\%\) confidence interval, and a test of whether the two laboratories are measuring the same thing.

    Solution

    The two sample means are sufficient: \(\bar X_1\sim N(\mu,\ 1/4)\) and \(\bar X_2\sim N(\mu,\ 4/16)=N(\mu,\ 1/4)\) , independent. The log-likelihood is \(-\tfrac12\cdot4(\bar x_1-\mu)^2-\tfrac12\cdot4(\bar x_2-\mu)^2\) plus a constant, maximised at the precision-weighted mean. Both precisions equal \(4\) , so

    \[ \hat\mu=\frac{4(10.4)+4(9.8)}{8}=10.1,\qquad\operatorname{Var}\hat\mu=\frac18,\qquad\text{standard error}\approx0.354 . \]

    Sixteen measurements with standard deviation \(2\) carry exactly the same information as four with standard deviation \(1\) . The \(95\%\) interval is \(10.1\pm1.960(0.354)=(9.41,\ 10.79)\) ; with known variances the normal pivot is exact.

    To test whether the laboratories agree, consider \(\bar X_1-\bar X_2\sim N(\mu_1-\mu_2,\ \tfrac14+\tfrac14)\) . Under \(H_0:\mu_1=\mu_2\) ,

    \[ z=\frac{10.4-9.8}{\sqrt{0.5}}\approx0.85,\qquad p\approx0.40 . \]

    There is no evidence of a discrepancy, so pooling is reasonable. The order of the steps matters: the combined estimate presumes a common \(\mu\) , and the test checks that presumption. Had the test rejected, the weighted mean would be an average of two different quantities and its confidence interval would be meaningless.

13. A random sample of size \(25\) from the density \(\theta x^{-\theta-1}\) on \(x>1\) has \(\sum\log x_i=10\) . Find the maximum likelihood estimate of \(\theta\) and its approximate standard error from the Fisher information, give the Wald \(95\%\) interval, and compare it with the exact interval obtained from the fact that \(\log X\) is exponential with rate \(\theta\) .

    Solution

    The log-likelihood is \(\ell(\theta)=n\log\theta-(\theta+1)\sum\log x_i\) , so \(\ell'(\theta)=n/\theta-\sum\log x_i\) and \(\ell''(\theta)=-n/\theta^2<0\) . The MLE is

    \[ \hat\theta=\frac{n}{\sum\log x_i}=\frac{25}{10}=2.5 . \]

    The second derivative does not involve the data, so \(I(\theta)=1/\theta^2\) and the approximate standard error is \(\hat\theta/\sqrt n=2.5/5=0.5\) . The Wald interval is \(2.5\pm1.960(0.5)=(1.52,\ 3.48)\) .

    For the exact interval, \(P(\log X>y)=P(X>e^y)=e^{-\theta y}\) , so \(\log X\sim\operatorname{Exp}(\theta)\) and, as in Example 2.4, \(2\theta\sum\log X_i\sim\chi^2_{50}\) . The \(2.5\%\) points of \(\chi^2_{50}\) are \(32.36\) and \(71.42\) , so

    \[ \frac{32.36}{20}\le\theta\le\frac{71.42}{20},\qquad\text{that is, }(1.62,\ 3.57). \]

    The Wald interval has almost the same length and is shifted to the left by about \(0.1\) : the sampling distribution of \(\hat\theta\) , a constant over a gamma variable, is skewed to the right, as in Figure 9.1, and the symmetric interval cannot reflect that. With \(n=25\) the discrepancy is modest. The example is Example 9.4 in disguise, since the transformation \(x\mapsto\log x\) turns this model into the exponential one; recognising a familiar model under a transformation is often the quickest route to an exact pivot.

14. An observational study of employees finds that those who took a training course earn more, with a regression coefficient that remains after adjusting for age and education. Explain what the coefficient does and does not establish, design a randomized experiment to estimate the effect of the course, write down the estimator and its variance estimate, and say which assumption of the observational analysis the experiment makes unnecessary.

    Hint

    The coefficient describes how mean earnings differ between employees who did and did not take the course, among those alike in age and education. By Theorem 10.8 it also absorbs every unmeasured influence on earnings that is associated with taking the course, such as motivation. For the experiment: take \(n\) willing employees, choose \(n_1\) by complete randomization to be offered the course, and compare mean earnings afterwards; the estimator is the difference in means, with variance estimate \(s_1^2/n_1+s_0^2/n_0\) (Theorems 12.4 and 12.5). The assumption made unnecessary is that no omitted variable is correlated with taking the course. Consider blocking on department or salary grade, and analyse by assignment if some decline the offer.

15. A normal sample of size \(16\) has mean \(50\) and standard deviation \(4\) . Compute the \(95\%\) confidence interval for the mean and the \(95\%\) prediction interval for a new observation, estimate the fraction of future observations that will fall inside the confidence interval, and say which of the two intervals should be judged by its coverage of future observations.

    Answer

    With \(t_{15,0.025}\approx2.131\) : confidence interval \(50\pm2.131(4/\sqrt{16})=(47.87,\ 52.13)\) ; prediction interval \(50\pm2.131(4)\sqrt{1+1/16}=50\pm8.79=(41.21,\ 58.79)\) . Using the estimates, a future observation falls in the confidence interval with probability about \(P(\lvert Z\rvert\le2.13/4)\approx0.41\) . Only the prediction interval makes a claim about future observations and should be judged by how often they fall inside it.

16. Let \(X=\mu+Z\) be a single observation, where \(Z\) has a known density \(f_0\) that is symmetric about zero and decreasing in \(\lvert z\rvert\) . Show that \(X\pm c\) , with \(c\) the upper \(\alpha/2\) point of \(f_0\) , is the shortest \(1-\alpha\) confidence interval of the form \([X-a,X+b]\) , and that under the flat prior \(\pi(\mu)\propto1\) the same interval is the highest posterior density credible interval of level \(1-\alpha\) .

    Solution

    Confidence interval. \(X-\mu=Z\) is a pivot with density \(f_0\) . For \(a,b\) with \(P(-b\le Z\le a)=1-\alpha\) , the event \(-b\le X-\mu\le a\) is the event \(X-a\le\mu\le X+b\) , so \([X-a,X+b]\) is a \(1-\alpha\) confidence interval, of length \(a+b\) . Every interval of the stated form arises this way. By Proposition 7.3, since \(f_0\) is symmetric about zero and decreasing in \(\lvert z\rvert\) , among intervals \([-b,a]\) with probability content \(1-\alpha\) under \(f_0\) the shortest is the symmetric one \([-c,c]\) with \(c\) the upper \(\alpha/2\) point. Hence \(X\pm c\) is the shortest.

    Credible interval. With \(\pi(\mu)\propto1\) , the posterior density is proportional to the likelihood: \(\pi(\mu\mid x)\propto f_0(x-\mu)\) . As a function of \(\mu\) this integrates to \(\int f_0(x-\mu)\,d\mu=1\) , so the posterior is proper and equals \(f_0(x-\mu)\) ; by symmetry of \(f_0\) it is the density \(f_0\) centred at \(x\) . A highest posterior density set is \(\{\mu:f_0(x-\mu)\ge k\}\) . Because \(f_0\) is decreasing in \(\lvert z\rvert\) , this is \(\{\mu:\lvert x-\mu\rvert\le c'\}\) for some \(c'\) , an interval symmetric about \(x\) . Its posterior probability is \(P(\lvert Z\rvert\le c')\) , which equals \(1-\alpha\) exactly when \(c'=c\) .

    So the two intervals coincide: \(x\pm c\) . The reason is the one met in Exercise 11.9. In a location model the pivot \(X-\mu\) has the same distribution whether \(X\) is regarded as random with \(\mu\) fixed, or \(\mu\) as random with \(X\) fixed under the flat prior. The frequentist seeks the shortest interval and the Bayesian the region of highest density, and for a symmetric unimodal density these are the same set. The coincidence fails as soon as the model is not a pure location family or the prior is not flat.